Question

A body is travelling with \(10\,\text{m s}^{-1}\) on a rough horizontal surface. Its velocity after \(2\,\text{s}\) is \(4\,\text{m s}^{-1}\). The coefficient of kinetic friction between the block and the plane is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))

• A. \(0.4\)
• B. \(0.3\)
• C. \(0.5\)
• D. \(0.2\)

Hint:

For motion on a horizontal rough surface, friction produces retardation: \[ a=\mu g \] Always first calculate acceleration using equations of motion and then compare with \(\mu g\).

Answer 1

The Correct Option is B

Step 1: Identify the known quantities.
Initial velocity: $u = 10$ m/s. Final velocity: $v = 4$ m/s. Time: $t = 2$ s. Acceleration due to gravity: $g = 10$ m/s^2. The block decelerates due to kinetic friction.
Step 2: Use the first equation of motion to find acceleration.
\[ v = u + at \] \[ 4 = 10 + a(2) \] \[ 2a = 4 - 10 = -6 \] \[ a = -3 \text{ m/s}^2 \] The negative sign indicates deceleration. The magnitude of retardation is $3$ m/s^2.
Step 3: Relate retardation to friction force.
On a horizontal surface, only friction acts to decelerate the block. Applying Newton's second law: \[ ma = -\mu_k mg \] The retardation equals $\mu_k g$: \[ \mu_k g = 3 \]
Step 4: Solve for the coefficient of kinetic friction.
\[ \mu_k = \frac{3}{g} = \frac{3}{10} = 0.3 \]
Step 5: Physical interpretation.
The friction force $f = \mu_k N = \mu_k mg$ acts opposite to the direction of motion. This friction caused a deceleration of 3 m/s^2, consistent with $\mu_k = 0.3$.
Step 6: State the final answer.
The coefficient of kinetic friction between the block and the plane is $0.3$. \[ \boxed{\mu_k = 0.3} \]